✔ 最佳答案
設A為(a,b) B為(p,q) (a>p)
y = –x2/2 + h
P = (2,4)在拋物線上,
所以 4 = –2 + h => h = 6
拋物線上任何兩點(m,n),(h,k),其斜率為
(k–n)/(h–m) = (–h2/2 + 6 + m2/2 –6)/(h–m) = (–1/2)(m+h)
PA斜率 = (–1/2)(a+2)
PB斜率 = (–1/2)(p+2)
PA, PB斜角互補即 (–1/2)(a+2) = (–1)(–1/2)(p+2)或a+p = –4
AB斜率 = (–1/2)(a+p) = 2 為定值
視三角底線為AB, 則高為P和AB的距離.
AB2 = (a–p)2 + (b–q)2
=(a–p)2 + (–a2/2 + 6 + p2/2 –6)2
=(a–p)2 +[1/2(p–a)(p+a)]2
=5(a–p)2
=20(a+2)2; 因 p = –4 – a
AB = 2√5(a+2)
AB方程式為 y=mx+c,已知m=2.
代入A點:b = 2a + c => c = b–2a = 6 – a2/2 – 2a
三角高 = [4 – (2)(2) + c]/ √(12+22) = (6 – a2/2 – 2a) /√5
三角面積 T = (1/2)[ 2√5(a+2)][ (6 – a2/2 – 2a) /√5
T = (a+2)( 6 – a2/2 – 2a)
dT/da = 6 – a2/2 – 2a + (a+2)(–a–2)
dT/da = (–3/2)a2 – 6a + 2
dT/da = 0 => a = –2 + 4√3/3 (p = –2 – 4√3/3)
d2T/da2 = –3a – 6 = –4√3/3 => 最大面積
三角最大面積 = (a+2)( 6 – a2/2 – 2a)
= (4√3/3)(6 – 2 + 8√3/3 – 8/3 + 4 – 8√3/3)
= (4√3/3)(8 – 8/3)
= 64√3/9
